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Research article

Understanding the oscillations of an epidemic due to vaccine hesitancy

  • Vaccine hesitancy threatens to reverse the progress in tackling vaccine-preventable diseases. We used an SIS model with a game theory model for vaccination and parameters from the COVID-19 pandemic to study how vaccine hesitancy impacts epidemic dynamics. The system showed three asymptotic behaviors: total rejection of vaccinations, complete acceptance, and oscillations. With increasing fear of infection, stable endemic states become periodic oscillations. Our results suggest that managing fear of infection relative to vaccination is vital to successful mass vaccinations.

    Citation: Anthony Morciglio, R. K. P. Zia, James M. Hyman, Yi Jiang. Understanding the oscillations of an epidemic due to vaccine hesitancy[J]. Mathematical Biosciences and Engineering, 2024, 21(8): 6829-6846. doi: 10.3934/mbe.2024299

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  • Vaccine hesitancy threatens to reverse the progress in tackling vaccine-preventable diseases. We used an SIS model with a game theory model for vaccination and parameters from the COVID-19 pandemic to study how vaccine hesitancy impacts epidemic dynamics. The system showed three asymptotic behaviors: total rejection of vaccinations, complete acceptance, and oscillations. With increasing fear of infection, stable endemic states become periodic oscillations. Our results suggest that managing fear of infection relative to vaccination is vital to successful mass vaccinations.



    Rotating blades (thin-walled beam) are important structures widely used in mechanical and aerospace engineering as aviation engine blades, various cooling fans, windmill blades, helicopter rotor blades, airplane propellers etc. The study of the dynamics of rotating blades is important to design purposes, optimization, and control.

    If the shear effect is not considered, the Euler-Bernoulli beam equation is used to model vibration of thin-walled beam. Chen et al.[6] studied the boundary feedback stabilization of a linear Euler-Bernoulli beam equation, they proved that the total energy of the equation decays uniformly and exponentially with t, when the beam is clamped at the left end and subjected to a feedback boundary conditions at the right end. Then Guo et al.[12] proved the well-posedness and stability of the system proposed by Chen et al.[14] considered a nonlinear Euler-Bernoulli beam equation with a feedback force applied at the free end, the existence of the weak and classical solutions were proved. Other relevant studies are referred to Refs. [4,18,19,17,13]. Note that the above literatures studied the longitudinal vibration in one direction.

    A very extensive work devoted to the longitudinal vibration in two directions were done by Librescu and Song[21,32,33] and their co-workers[25,28]. Under assumption of the cross-section to be rigid in its own plane, they modelled the rotating blades by 1-D linear governing equations. The influence of many factors on rotating blades, such as the anisotropy and heterogeneity of constituent materials, functionally graded materials (FGM), temperature, shear effects, primary and secondary warping phenomena (Vlasov effect), centrifugal and Coriolis forces etc have been taken into account in the 1-D linear governing equations.

    Following Librescu's approach, various blades models were derived. Georgiades et al.[11] modelled a rotating blades by means of linear strain-displacement relationships, considering arbitrary pitch (presetting) angle and non-constant rotating speed. Choi et al.[8] studied bending vibration control of the pre-twisted rotating composite blades, who emphasized the important of piezoelectric effect in single cell composite blades. Fazelzadeh et al.[9] considered a thin-walled blades made of FGM which is used in turbomachinery under aerothermodynamics loading. In the paper, quasi-steady aerodynamic pressure loadings was determined by the first-order piston theory, and steady beam surface temperature was obtained from gas dynamics theory. Fazelzadeh et al.[10] studied the governing equations which included the effects of the presetting angle and the rotary inertia. The effects of steady wall temperature and quasi-steady aerodynamic pressure loadings due to flow motion were also taken into account.

    The models in Refs.[21,32,33,25,28,11,8,9,10] are linear. When the engine blades rotate at a low speed, the linear approximation can completely meet the needs of practical application. However, when the blade rotate at a high speed, the simple linear approximation can not accurately describe the dynamic behavior of the system. So the non-linear analysis of rotating blades has attracted considerable attention.

    The nonlinear governing equations of a rotating blades at constant angular velocity was presented by Anderson[1], and the author linearized the equation under the assumption that a small perturbed motion occurred at an initially stressed equilibrium configuration. Chen et al.[7] considered the effects of geometric non-linearity, shear deformation and rotary inertia. Arvin et al.[2] builded a nonlinear governing equation for rotating blades considering centrifugal forces by means of von-Karmans strain-displacement relationships under assumption of the constant rotation speed and zero pitch (presetting) angle. Yao et al.[39] employed the Hamilton's principle to derive the nonlinear governing equations with periodic rotating speed, arbitrary pitch (presetting) angle and linear pre-twist angle. Under the assumption that the location of shear centre is different from the centre of gravity, Avramov et al.[3] obtained results of the investigations on flexural-flexural-torsional nonlinear vibrations of twisted rotating blades described by the model of three nonlinear integro-differential equations. Other nonlinear models can be found in Refs.[34,29,27,37,15,31,36].

    To the best of the author's knowledge, all the above literatures about the longitudinal vibration in two directions skipped the existence proof of solutions to the governing equations, and directly used the finite element method to study the influence of various parameters on blades vibration. To address this situation, we first try to model a governing equations of the blades with arbitrary rotating speed, arbitrary pre-setting angle and pre-twist angle. In the process of building the model, we take into account the free vibration at the right end of the blade and the nonlinear relationship between stress and strain. In the paper we aim to investigate the well-posedness and regularity of the governing equations. The well-posedness of other nonlinear blade vibration models can be found in Refs. [5,38,26,35,16].

    Let us consider a slender, straight blades mounted on a rigid hub of radius R0 rotating with the angular velocity ω(t). The length of the blades is denoted by l, its wall thickness by h, the length and the width of the cross section of the blades are a and b, respectively. The sum of pre-twist angle α(x) and pre-setting angle β is defined by θ(x), as shown in Figure 1 and 2.

    Figure 1.  Rotating blades.
    Figure 2.  The cross-section of rotating blades.

    To derive the model of the rotating blades, the following kinematic and static assumptions are postulated:

    (ⅰ) The blades is perfectly elastic bodies, the blades material is isotropic and is not affected by temperature,

    (ⅱ) The cross section of the blades is rectangular and all its geometrical dimensions remain invariant in its plane,

    (ⅲ) The ratio of wall thickness h to the radius of curvature r at any point of the blade wall is negligibly small while compared to unity,

    (ⅳ) The transverse shear effect of the cross section is neglected,

    (ⅴ) The axial displacement w is much smaller than u or v and the derivatives of w can be neglected in the strain-displacement relations. where u,v,w represent the displacement of the middle line of cross-sections along the y,z,x axis, respectively.

    Inertial Cartesian coordinate systems (X,Y,Z) attached to the center of the hub O, the unit vectors of the inertial coordinate systems (X,Y,Z) is defined as (I,J,K).

    Rotating coordinate systems (x,y,z) located at the blade root, the origin o of the rotating coordinate systems is set at the center of the beam cross section, the unit vectors of the inertial coordinate systems (x,y,z) is defined as (i,j,k).

    Transformation between XYZ and xyz frames can be written in the form

    (X,Y,Z)=(x,y,z)+(R0,0,0). (1)

    Local coordinates systems (xp,yp,zp) also located at the blade root and oriented with respect to plane of rotation (y,z) at angle ω(t). So (x,y,z)=B(xp,yp,zp), where the rotation matrix B is

    (1000cosωsinω0sinωcosω).

    Local, curvilinear coordinate systems (x,n,s) related to blade cross section (Figure 2). Its origin is set conveniently at the point on a mid-line contour. s and n are the circumferential and thickness coordinate variables, the unit vectors of axis n and s are defined as en,et, respectively.

    In order to determine the relationship between the two coordinate systems (x,y,z) and (x,n,s), one defines the position vector r(r(s,x)) from the reference x-axis of the blades to an arbitrary point A located on the middle surface as

    r=xi+y(s)j+z(s)k. (2)

    The position vector r of an arbitrary point A off the mid-surface of the blades can be expressed as

    r=r+nen, (3)

    As a result

    et=drds=dy(s)dsj+dz(s)dsk, (4)
    en=et×i=dz(s)dsjdy(s)dsk. (5)

    Moreover, in order to avoid confusion, the notation (x,y,z) represents the points associated with the middle surface, the notation A=(x,y,z) represents the points off the middle surface, the two notations (x,y,z) and (x,y,z) are presented as follows:

    x=x,  y=rj=y+ndzds,  z=rk=zndyds. (6)

    Based on the assumptions (ⅳ) and (ⅴ), the axial displacement Dx,Dy,Dz (see Figure 3) are expressed as (see Ref. [21])

    Dx=ϕy(x,t)(z(s)ndyds)+ϕz(x,t)(y(s)+ndzds),Dy=u,Dz=v. (7)
    Figure 3.  Position vectors of ponit A in reference frames.

    where ϕy,ϕz denote rotation of the cross-section about y and z axis. For non-shearable blades, The expression of ϕy and ϕz can be written in the form

    ϕy=vx,ϕz=ux. (8)

    where ux,vx denote derivatives with respect to x.

    Based on the assumptions (ⅴ), the displacement-strain relationships is expressed as follows[20]:

    εxx=Dxx+12[(Dyx)2+(Dzx)2]. (9)

    Thanks to (8), we can get

    εxx=ˉεxx+ˉˉεxxn,

    where

    ˉεxx=12(u2x+v2x)(uxxy(s)+vxxz(s)),  ˉˉεxx=uxxdzds+vxxdyds. (10)

    The shear strain εnx can be expressed as

    εnx=12(Unx+wn), (11)

    where Un represents the components of (Dx,Dy,Dz) along the n axis,

    Un=(Dx,Dy,Dz)en=udzdsvdyds. (12)

    Substituting (7) and (12) into (11), we deduce εnx=0. Using the coordinate transformation of strain component, taking into account assumptions (iv), the shear strain εsx can be expressed as

    εsx=12(dydsγxy+dzdsγxz)=0. (13)

    where γxy,γxz are the transverse shear effect of the cross section.

    Since these materials are isotropic, the corresponding thermoelastic constitutive law adapted to the case of structures is expressed as

    (σssσxxσxnσnsσsx)=(Q11Q12000Q21Q2200000Q4400000Q5500000Q66)(εssεxxεxnεnsεsx),

    Herein, the reduced thermoelastic coefficients are defined as:

    Q11=Q22=E1ν2,Q12=Q21=Eν1ν2,Q44=Q55=k2E2(1+ν),Q66=E2(1+ν),

    where E is Young's modulus, ν is Poisson's ratio, k2 is the transverse shear correction factor.

    According to assumption (ⅱ), the cross section is rigid, then we can derive σns=0. Considering the assumption of the hoop stress σss to be negligible, we can get

    (σss,σxx,σxn,σns,σsx)=(0,Eεxx,0,0,0).

    The centrifugal force can be represented as

    Fc=lxρˉAω2(R0+x)dx=ρˉAω2R(x),

    where ρ is the density of the blades, ˉA is the cross section area of the blades, R(x)=R0(lx)+12(l2x2).

    In the paper, we use the first-order piston theory (see Ref.[24]) to evaluate the perturbed gas pressure. The pressure on the principal plane of the blade can be obtained as

    Pyp=Cρ(upt+Utypupx),Pzp=Cρ(vpt+Utzpvpx), (14)

    where

    Utyp=Ucosθ,  Utzp=Usinθ, (15)

    C represents the speed of sound, ρ and U, respectively, denote the density and the velocity of the free stream air, Utyp and Utzp are, respectively, the tangential components of the fluid velocity on the positive yp and zp plane, and up and vp denote the displacement components along the principal axes yp and zp, respectively.

    The transformation relationship between (up,vp) and (u,v) is given by

    up=ucosθ+vsinθ, vp=usinθ+vcosθ. (16)

    Therefore, the external forces per unit axial length in the y direction and the z direction can be obtained as

    py=aPzpsinθbPypcosθ, pz=aPzpcosθbPypsinθ. (17)

    Combining (14), (15), (16), py,pz can be expressed as a linear function of u,v,ut,vt,ux,vx, respectively.

    py=b1ux+b2vx+b3u+b4v+b5ut+b6vt (18)
    pz=e1ux+e2vx+e3u+e4v+e5ut+e6vt (19)

    where

    b1=CρU(asin3θ+bcos3θ)b2=CρUsinθcosθ(asinθ+bcosθ)b3=CρUθxsinθcosθ(asinθbcosθ)b4=CρUθx(asin3θ+bcos3θ)b5=Cρ(asin2θ+bcos2θ)b6=Cρsinθcosθ(ab)e1=CρUsinθcosθ(asinθ+bcosθ)e2=CρUsinθcosθ(acosθ+bsinθ)e3=CρUθxsinθcosθ(asinθ+bcosθ)e4=CρUθxsinθcosθ(asinθ+bcosθ)
    e5=Cρsinθcosθ(ab)e6=Cρ(acos2θ+bsin2θ)

    In order to calculate the kinetic energy, the velocity vector and the acceleration should be given first. Based on the assumption (ⅴ), the position vector R of a point A of the deformed blades is expressed in the form

    R(X,Y,Z,t)=xi+(y+u)j+(z+v)k+R0i.

    Keep in mind that the rotation takes place solely in the XY plan, it results

    it=ωj;  jt=ωi;  kt=0.

    Then the velocity vector and the acceleration of an arbitrary point A are obtained:

    Rt=ω(y+u)i+(ω(R0+x)+ut)j+vtk, (20)
    Rtt=[2ωut+ωt(y+u)+ω2(R0+x)]i         +[uttωt(R0+x)ω2(y+u)]j+vttk, (21)

    where subscript t denotes derivatives with respect to the time.

    In order to derive the blades model and the associated boundary conditions, the extended Hamilton's principle is used. This can be formulated as

    t2t1(δKδU+δW)dt=0,δu=0,δv=0att=t1,t2. (22)

    where K and U, respectively, denote the kinetic energy and the strain energy, W is the virtual work of external forces, δ is the variation operator.

    Thanks to the cross section of the blades is rectangular, we get

    y(s)ds=z(s)ds=0. (23)

    Utilizing (23), the kinetic energy is obtained

    K=12τρR2tdτ=12ρτu2t+v2t+2ω(R0+x)ut        +ω2(u2+2(y+ndzds)u+y+ndzds+(R0+x)2)dτ=12ρˉAl0u2t+v2t+2ω(R0+x)ut        +ω2(u2+2hydsˉAu+hydsˉA+(R0+x)2)dx=12ρˉAl0u2t+v2t+2ω(R0+x)ut+ω2(u2+(R0+x)2)dx,

    where dτ denotes the differential volume element. Then

    t2t1δKdt=ρˉAt2t1l0{[utt+ωt(R0+x)uω2]δu+vttδv}dxdt, (24)

    Due to the rotating motion of the blades, the total strain energy consists of two parts. The strain energy caused by the centrifugal force can be obtained as

    U1=12l0h2h2lxρω2(R0+ς)εxxdςdndsdx (25)

    where ρ is the bulk density of the blades, lxρω2(R0+ς)dς is the centrifugal force per unit cross section. Then,

    δU1=12l0h2h2lxρω2(R0+ς)δεxxdςdndsdx=12l0h2h2lxρω2(R0+ς)δˉεxxdςdndsdx=ρˉAω2l0R(x)(uxδux+vxδvx)dxρω2h2l0lx(R0+ς)(y(s)dsδuxx+z(s)dsδvxx)dςdx, (26)

    where ˉA is the area of the cross section of the blades, R(x)=R0(lx)+12(l2x2).

    Thanks to (23), the variation of the strain energy caused by the centrifugal force can be rewritten as

    δU1=ρˉAω2[R(x)(uxδu+vxδv)]|l0l0(R(x)ux)xδu+(R(x)vx)xδvdx (27)

    The strain energy induced by the deformation of the rotating blades can be expressed as

    U2=12l0h2h2σxxεxxdndsdx. (28)

    Substituting the expressions of the stress and strain resultants into (28) yields

    δU2=El0h2h2εxxδεxxdndsdx=Eh4l0(u2x+v2x)δ(u2x+v2x)dsdxEh2l0(u2x+v2x)δ(uxxy(s)+vxxz(s))dsdxEh2l0(uxxy(s)+vxxz(s))δ(u2x+v2x)dsdx+Ehl0(uxxy(s)+vxxz(s))δ(uxxy(s)+vxxz(s))dsdx+Eh312l0(uxxdzds+vxxdyds)δ(uxxdzds+vxxdyds)dsdx (29)
    δU2=El0h2h2εxxδεxxdndsdx=ρˉA{l0[a52((u2x+v2x)ux)x+(a6uxxa3vxx)xx          a12(u2x+v2x)xx+(a1uxx+a2vxx)ux)x]dx}δu+ρˉA{l0[a52((u2x+v2x)vx)x+(a4vxxa3uxx)xx          a22(u2x+v2x)xx+(a1uxx+a2vxx)vx)x]dx}δv+ρˉA{[(a6uxxa3vxx)a12(u2x+v2x)]δux}|l0+ρˉA{[(a4vxxa3uxx)a22(u2x+v2x)]δvx}|l0+ρˉA{[a52(u2x+v2x)ux(a6uxxa3vxx)x          +a12(u2x+v2x)x+a1uxx+a2vxxux]δu}|l0+ρˉA{[a52(u2x+v2x)vx(a4vxxa3uxx)x          +a22(u2x+v2x)x+a1uxx+a2vxxvx]δv}|l0 (30)

    where

    a1=EhˉAρyds=0;a2=EhˉAρzds=0;a3(x)=EhˉAρh212dydsdzdsyzds;a4(x)=EhˉAρh212(dzds)2+y2ds;a5=Eρa6(x)=EhˉAρh212(dyds)2+z2ds.

    The work of the non-conservative external forces can be obtained as

    W=l0pyu+pzvdx. (31)

    Then

    t2t1δWdt=t2t1l0˜pyδu+˜pzδvdxdt+t2t1{(b1u+e1v)δu+(b2u+e2v)δv}|l0dt. (32)

    where

    ˜py=2b3u+(e3+b4)v (33)
    ˜pz=(e3+b4)u+2e4v (34)

    Inserting variation of potential energy (27) and (30), variation of kinetic energy (24) and variation of external work equation (32) into the extended Hamilton's principle (22), collecting the terms associated with the same variations, invoking the stationarity of the functional within the time interval [t0,t1], and the fact that the variations (δu,δv) are independent and arbitrary, their coefficients in the two integrands must vanish independently [see [21]]. The partial differential equations with respect to variation of problem's independent variables and the associated boundary conditions are obtained as

    utt+(a6uxxa3vxx)xxa52((u2x+v2x)ux)x        ω2(R(x)ux)xω2u¯p1+ωt(R0+x)=0, (35)
    vtt+(a4vxxa3uxx)xxa52((u2x+v2x)vx)xω2(R(x)vx)x¯p2=0, (36)

    where

    ¯p1=˜pyρˉA,  ¯p2=˜pzρˉA. (37)

    The following two types of boundary conditions are generated due to the different design of engine blades.

    u,ux=0,v,vx=0,x=0, (38)
    u,ux=0,v,vx=0,x=l. (39)

    and

    u,ux=0,v,vx=0,x=0, (40)
    a6uxxa3vxx=0,x=l, (41)
    a4vxxa3uxx=0,x=l, (42)
    a52(u2x+v2x)ux(a6uxxa3vxx)xb1ue1v=0,x=l, (43)
    a52(u2x+v2x)vx(a4vxxa3uxx)xb2ue2v=0,x=l. (44)

    The conditions (38)-(39) represent C-C boundary condition. The conditions (40)-(44) represent C-F boundary condition. Now we study the well-posedness and regularity of the solution for C-C and C-F blades.

    We write Q=Ω×(0,T), where Ω=(0,l) and T>0,

    H20={ψH2(Ω)|ψ(x)=0,ψx(x)=0,xΩ},H2f={ψH2(Ω)|ψ(0)=0,ψx(0)=0,x=0}.

    We list Gagliardo-Nirenberg inequality for bounded domains (see [23]) to be used in the subsequent sections

    Lemma 3.1. Let ΩRn be a bounded domain with smooth boundary. Let 1p,q,r be real numbers and jm be non-negative integers. If a real number α satisfies

    1pjn=α(1rmn)+(1α)1q,jmα1

    Then

    DjfLp(Ω)C1DmfαLr(Ω)f1αLq(Ω)+C2fLs(Ω)

    where s>0, and the constants C1 and C2 depend upon Ω and the indices p,q,r,m,j,s only.

    Now, we give the Aubin-Lions Lemma (see [22]).

    Lemma 3.2. Suppose B0,B,B1 are Banach Space, if

    (i) {ui}i=1 is bounded in Lp0(0,T;B0);

    (ii) {ui,t}i=1 is bounded in Lp1(0,T;B1);

    (iii) B0↪↪BB1,

    then {ui}i=1 admits a strongly converging subsequence in Lp0(0,T;B), provided p0<,p1>1.

    Without loss of generality, we assume a5=1. For the sake of brevity, the initial boundary-value problems of C-C blades are rewritten as:

    {utt+(a6uxx)xx(a3vxx)xx12((u2x+v2x)ux)x    ω2(Rux)xp1=0inQ,vtt(a3uxx)xx+(a4vxx)xx12((u2x+v2x)vx)x    ω2(Rvx)xp2=0inQ,u,ux=0,v,vx=0onΩ×[0,T],u=u0(x),ut=u1(x),v=v0(x),vt=v1(x)onΩ×{t=0}. (45)

    where

    p1=¯p1+ω2uωt(R0+x),  p2=¯p2. (46)

    Definition 4.1. We say function (u,v), u,vL(0,T;H20(Ω)), with

    ut,vtL(0,T;L2(Ω)),utt,vttL(0,T;H2(Ω)),

    is a weak solution of the initial boundary value problem (45) provided

    (ⅰ)

    (utt,φ)+(a6uxx,φxx)(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)(p1,φ)=0. (47)
    (vtt,φ)(a3uxx,φxx)+(a4vxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)(p2,φ)=0. (48)

    for each φH20(Ω), and a.e. time 0tT, and

    (ⅱ)

    u(0)=u0,ut(0)=u1;v(0)=v0,vt(0)=v1. (49)

    Remark 1. From the definition, we know u,vC([0,T];L2(Ω)) and ut,vtC([0,T];H2(Ω)).

    Remark 2. By the product Minkowshi inequality, we can obtain:

    a4a6a23>0,xΩ. (50)

    Theorem 4.2. (Existence for weak solution of (45)) Assume

    ωC1(0,T),a3,a4,a6L(Ω),u0,v0H20(Ω),u1,v1L2(Ω). (51)

    there exists a weak solution of (45).

    We now briefly outline the proof of Theorem 4.2 in the following:

    Step 1. employing Galerkin's method to construct solutions of certain finite-dimensional approximations to (45) (correspond to Lemma 6.1 in chapter 6);

    Step 2. using the energy method to find the uniform estimates of the finite-dimensional approximations solutions (correspond to Lemma 6.2 in chapter 6);

    Step 3. using compactness method to obtain the weak solutions of (45).

    Now we give the smoothness of weak solutions of (45).

    Theorem 4.3. (Improved regularity) Assume

    {a3,a4,a6L(Ω),ωC1(0,T),ωttL(0,T),u0H20(Ω)H4(Ω),v0H20(Ω)H4(Ω),u1H20(Ω),v1H20(Ω), (52)

    the weak solution of (45) satisfies

    ut,vtL(0,T;H20(Ω)),utt,vttL(0,T;L2(Ω)). (53)

    Theorem 4.4. (Interior regularity) Under the condition (52), for any φ(x)C0(Ω),φ(x)>0,xΩ, then the weak solution of (45) satisfies

    φuxxx,φvxxxL(0,T;L2(Ω)).

    Remark 3. Multiplying the first and second equation of (45) by a4(x)utt,a3(x)vtt respectively, integrating with over Ω, and summing the two equations, we can obtain that the weak solution of (45) satisfies

    uxxxx,vxxxxL(0,T;L2(Ω)).

    where Ω is a bounded open interval and ¯ΩΩ.

    If the pre-twist angle α(x) is neglected, we can get θ(x),a3(x),a4(x),a6(x) are all constant. Then we give the following theorem.

    Theorem 4.5. (Improved regularity when α(x)=0) Under the condition (52), assume α(x)=0, then the weak solution of (45) satisfies

    u,vL(0,T;H4(Ω)H20(Ω)).

    Now, we study the uniqueness and stability of (45), denote operator π satisfying

    π:{a3,a4,a6,ω,θ,u0,v0,u1,v1}{u,v},

    and

    W(Q)={ψ|ψL(0,T;H20(Ω)),ψtL(0,T;L2(Ω))}

    with the norm

    ||ψ||W(Q)=||ψ||L(0,T;H20(Ω))+||ψt||L(0,T;L2(Ω)).

    Then W(Q) is Banach space.

    Theorem 4.6. Undering the condition (52),

    π:{(L(Ω))3×W1,1(0,T)C1(0,T)×L1(Ω)×(H20(Ω))2×(L2(Ω))2}(W(Q))2

    is continuous.

    The initial boundary-value problems of C-F blades are rewritten as :

    {utt+(a6uxx)xx(a3vxx)xx12((u2x+v2x)ux)x    ω2(Rux)xp1=0inQ,vtt(a3uxx)xx+(a4vxx)xx12((u2x+v2x)vx)x    ω2(Rvx)xp2=0inQ,u,ux=0,v,vx=0,x=0,a6uxxa3vxx=0,x=l,a4vxxa3uxx=0,x=l,12(u2x+v2x)ux(a6uxxa3vxx)xb1ue1v=0,x=l,12(u2x+v2x)vx(a4vxxa3uxx)xb2ue2v=0,x=l,u=u0(x),ut=u1(x),v=v0(x),vt=v1(x)onΩ×{t=0}. (54)

    Definition 5.1. We say function (u,v), u,vL(0,T;H2f(Ω)), with

    ut,vtL(0,T;L2(Ω)),utt,vttL(0,T;H2(Ω)),

    is a weak solution of the initial boundary value problem (54) provided

    (ⅰ)

    (utt,φ)+(a6uxx,φxx)(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)(p1,φ)(b1u(l)+e1v(l))φ(l)=0. (55)
    (vtt,φ)(a3uxx,φxx)+(a4vxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)(p2,φ)(b2u(l)+e2v(l))φ(l)=0. (56)

    for each φH2f(Ω), and a.e. time 0tT, and

    (ⅱ)

    u(0)=u0,ut(0)=u1;v(0)=v0,vt(0)=v1.

    Similar as above, we derive the following conclusion of the initial boundary value problem (54).

    Theorem 5.2. (Existence for the weak solutions of (54)) Assume

    ωC1(0,T),a3,a4,a6L(Ω),u0,v0H2f(Ω),u1,v1L2(Ω), (57)

    there exists a weak solution of (54).

    Theorem 5.3. (Regularity weak solution of (54)) Assume

    {a3,a4,a6L(Ω),ωC1(0,T),ωttL(0,T),u0,v0H2f(Ω)H4(Ω),u1,v1H2f(Ω). (58)

    Then there exists T<T such that, the weak solution of (54) satisfies

    ut,vtL(0,T;H2f(Ω)),utt,vttL(0,T;L2(Ω)). (59)

    Remark 4. If pre-twist angle α(x) is neglected, according to the first and second equations of (54), we can obtain that the weak solution of (54) satisfies

    u,vL(0,T;H4(Ω)H2f(Ω)). (60)

    where utt,vttL(0,T;L2(Ω)) and Gagliardo-Nirenberg inequality are used.

    We construct weak solution of the initial boundary-value problem (45) by first solving a finite dimensional approximation. We thus employ Galerkin's method by selecting smooth functions φk=φk(x)(k=1,) such that

    {φk}k=1is an orthogonal basis ofH20(Ω), (61)

    and

    {φk}k=1is an orthonormal basis ofL2(Ω). (62)

    Fix a positive integer m, and write

    um=mk=1dk0m(t)φk, vm=mk=1dk1m(t)φk, (63)

    where we intend to select the coefficients dk0m(t),dk1m(t)(0tT,k=1,,m) to satisfy

    dk0m(0)=(u0,φk),dk1m(0)=(v0,φk),k=1,,m, (64)
    dk0m,t(0)=(u1,φk),dk1m,t(0)=(v1,φk),k=1,,m, (65)

    and

    (um,tt,φk)+(a6um,xx,φk,xx)(a3vm,xx,φk,xx)+12((u2m,x+v2m,x)um,x,φk,x)+ω2(Rum,x,φk,x)(p1m,φk)=0, (66)
    (vm,tt,φk)(a3um,xx,φk,xx)+(a4vm,xx,φk,xx)+12((u2m,x+v2m,x)vm,x,φk,x)+ω2(Rvm,x,φk,x)(p2m,φk)=0. (67)

    where

    p1m=1ρˉA(2b3um+(e3+b4)vm)+ω2umωt(R0+x),p2m=1ρˉA((e3+b4)um+2e4vm).

    In order to proof Theorem 4.2, we need the following Lemma.

    Lemma 6.1. Under the condition (51), for each integer m=1,2,, there exists a unique function um,vm of the form (63) satisfying (64)-(67) for 0ttm.

    Proof. Assuming um,vm to be given by (63), by using (62), equation (66) and (67) become the nonlinear system of ODE

    dk0m,tt+mj=1dj0m(a6φj,xx,φk,xx)mj=1dj1m(a3φj,xx,φk,xx)+12mj=1dj0m(((mi=1di0mφi,x)2+(mi=1di1mφi,x)2)φj,x,φk,x)+ω2mj=1dj0m(Rφj,x,φk,x)1ρˉA(2b3dk0m+(e3+e4)dk1m)=ω2dk0mωt((R0+x),φk), (68)
    dk1m,tt+mj=1dj1m(a4φj,xx,φk,xx)mj=1dj0m(a3φj,xx,φk,xx)+12mj=1dj1m(((mi=1di0mφi,x)2+(mi=1di1mφi,x)2)φj,x,φk,x)+ω2mj=1dj1m(Rφj,x,φk,x)1ρˉA((e3+e4)dk0m+2e4dk1m)=0. (69)

    subject to the initial conditions (64), (65). According to standard theory for ODE, there exists unique C2 funcion d0m(t)=(d10m,d20m,,dm0m),d1m(t)=(d11m,d21m,,dm1m), satisfying (64), (65), and solving (68), (69) for 0ttm, where tm:=t(m) is a function of m.

    We propose now to send m to infinity and to show a subsequence of the solutions um,vm of the approximate problem (64)-(67) converges to a weak solution of (45). For this we will need the following uniform estimates.

    Lemma 6.2. Undering the condition (51), there exists positive constant C(Ω,T), such that

    ||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω)C(||u1||2L2(Ω)+||v1||2L2(Ω)+||u0||2H20(Ω)+||v0||2H20(Ω))+C (70)

    for m=1,2,.

    Proof. Multiplying equality (66) by dk0m,t, summing k=1,2,, we deduce

    (um,tt,um,t)+(a6um,xx,um,txx)(a3vm,xx,um,txx)+12((u2m,x+v2m,x)um,x,um,tx)+ω2(Rum,x,um,tx)(p1m,um,t)=0 (71)

    for a.e. 0ttm.

    Multiplying equality (67) by dk1m,t, summing k=1,2,, we deduce

    (vm,tt,vm,t)(a3um,xx,vm,txx)+(a4vm,xx,vm,txx)+12((u2m,x+v2m,x)vm,x,vm,tx)+ω2(Rvm,x,vm,tx)(p2m,vm,t)=0 (72)

    for a.e. 0ttm.

    To simplify the equation (71) and (72), we can get

    12ddtum,t2L2(Ω)+12ddta6um,xx2L2(Ω)(a3vm,xx,um,txx)+12((u2m,x+v2m,x)um,x,um,tx)+ω22ddtRum,x2L2(Ω)(p1m,um,t)=0. (73)
    12ddtvm,t2L2(Ω)+12ddta4vm,xx2L2(Ω)(a3um,xx,vm,txx)+12((u2m,x+v2m,x)vm,x,vm,tx)+ω22ddtRvm,x2L2(Ω)(p2m,vm,t)=0. (74)

    Summing the equations (73) and (74), we discover

    12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||a6um,xx||2L2(Ω)+||a4vm,xx||2L2(Ω)+ω22(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))}ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)=ωωt(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))+(p1m,um,t)+(p2m,vm,t). (75)

    Since ωC1(0,T), the equality (75) implies

    12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||a6um,xx||2L2(Ω)+||a4vm,xx||2L2(Ω)+ω22(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))}ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)C(||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H10(Ω)+||vm||2H10(Ω))+C. (76)

    where we used Young inequality.

    Integrating (76) with respect to t, we discover

    ||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||a6um,xx||2L2(Ω)+||a4vm,xx||2L2(Ω)2||a3um,xxvm,xx||L1(Ω)+ω22(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))+14||u2m,x+v2m,x||2L2(Ω)C(Ω,T){||u1m||2L2(Ω)+||v1m||2L2(Ω)+||u0m||2H2(Ω)+||v0m||2H2(Ω)}+Ct0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H10(Ω)+||vm||2H10(Ω)+C. (77)

    Thanks to (50), there exists a constant C, such that

    ||a6um,xx||2L2(Ω)+||a4vm,xx||2L2(Ω)2||a3um,xxvm,xx||L(Ω)C(||um||2H20(Ω)+||vm||2H20(Ω)). (78)

    Substituting (78) into the inequality (77), By using Poincaré inequality, we find

    ||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω) C(||u1m||2L2(Ω)+||v1m||2L2(Ω)+||u0m||2H20(Ω)+||v0m||2H20(Ω))+Ct0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H20(Ω)+||vm||2H20(Ω)+C. (79)

    Then, by using Gronwall inequality, we obtain

    ||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω)C(||um,t(0)||2L2(Ω)+||vm,t(0)||2L2(Ω)+||um(0)||2H20(Ω)+||vm(0)||2H20(Ω))+CC(||u1||2L2(Ω)+||v1||2L2(Ω)+||u0||2H20(Ω)+||v0||2H20(Ω))+C

    for m=1,2,.

    Remark 5. From the Energy estimates, we can obtain: tmT, as m.

    Thanks to Lemma 6.1 and Lemma 6.2, we can obtain the existence for the weak solutions of the initial boundary value problem (45).

    Proof. (ⅰ) According to the energy estimates (70), we see that

    {um}m=1,{vm}m=1isboundedinL(0,T;H20(Ω)); (80)
    {um,t}m=1,{vm,t}m=1isboundedinL(0,T;L2(Ω)); (81)
    {um,tt}m=1,{vm,tt}m=1isboundedinL(0,T;H2(Ω)). (82)

    As a consequence there exists subsequence {uμ,vμ}m=1 and u,vL(0,T;H20(Ω)), with ut,vtL(0,T;L2(Ω)), utt,vttL(0,T;H2(Ω)), such that

    {uμu,vμvweaklyin  L(0,T;H20(Ω))uμ,tut,vμ,tvtweaklyin  L(0,T;L2(Ω))uμ,ttutt,vμ,ttvttweaklyin  L(0,T;H2(Ω)). (83)

    (ⅱ) By Gagliardo-Nirenberg inequality, we can see

    ||uμ,x||L(Q)C||uμ,xx||L(0,T;L2(Ω))C. (84)

    Otherwise,

    ||u2μ,x+v2μ,x||L(0,T;L2(Ω))||uμ,x||2L(0,T;L4(Ω))+||vμ,x||2L(0,T;L4(Ω))C||uμ||2L(0,T;H20(Ω))+c||vμ||2L(0,T;H20(Ω))C. (85)

    Combining (84) and (85), we discover

    ||(u2μ,x+v2μ,x)uμ,x||L(0,T;L2(Ω))C||u2μ,x+v2μ,x||L(0,T;L2(Ω))C. (86)

    Moreover, there exists a function χL(0,T;L2(Ω)) satisfy

    (u2μ,x+v2μ,x)uμ,xχ  weakly * in  L(0,T;L2(Ω)). (87)

    By Lemma 3.2, we can find

    uμu,vμv  strongly in  L2(0,T;H10(Ω)). (88)

    And so

    uμ,xux,vμ,xvx  strongly in  L2(Q), (89)

    Thus

    (u2μ,x+v2μ,x)uμ,x(u2x+v2x)ux. (90)

    Combining (87) and (90), we can obtain

    (u2μ,x+v2μ,x)uμ,x(u2x+v2x)ux  weakly in  L2(Q), (91)

    where we used the Lemma 1.3 of Chapter 1 in [22]. Furthermore, we have χ=(u2x+v2x)ux.

    Meanwhile,

    ((u2μ,x+v2μ,x)uμ,x,φk,x)((u2x+v2x)ux,φk,x)  weakly * in  L(0,T). (92)

    In the same way,

    ((u2μ,x+v2μ,x)vμ,x,φk,x)((u2x+v2x)vx,φk,x)  weakly * in  L(0,T). (93)

    Next fix an integer k, we select μk, from (66) we can get

    (uμ,tt,φk)+(a6uμ,xx,φk,xx)(a3vμ,xx,φk,xx)+12((u2μ,x+v2μ,x)uμ,x,φk,x)ω2(Ruμ,x,φk,x)(p1μ,φk)=0. (94)

    Thanks to (83), we can get

    {(uμ,tt,φk)(utt,φk)weaklyinL(0,T),(a6uμ,xx,φk,xx)(a6uxx,φk,xx)weaklyinL(0,T),(a3vμ,xx,φk,xx)(a3vxx,φk,xx)weaklyinL(0,T),(p1μ,φk)(p1,φk)weaklyinL(0,T). (95)

    From (92) and (95), we can discover

    +v2x)ux,φk,x)+ω2(Rux,φk,x)(p1,φk)=0 (96)

    for all fixed k.

    Note

    {φk}k=1  isanorthogonalbasisofH20(Ω),

    then,

    (utt,φ)+(a6uxx,φxx)(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)(p1,φ)=0 (97)

    for arbitrary φH20(Ω).

    In the same way,

    (vtt,φ)+(a4vxx,φxx)(a3uxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)(p2,φ)=0 (98)

    for arbitrary φH20(Ω).

    Synthesizes the above analysis, there exists u,v satisfying (47), (48), and

    u,vL(0,T;H20(Ω)),ut,vtL(0,T;L2(Ω)),utt,vttL(0,T;H2(Ω)).

    (ⅲ) Now let's prove the initial conditions.

    Since, uμ,vμ,uμ,t,vμ,t are bounded in L2(Q), by Lemma 1.2 of Chapter 1 in [22], we can discover

    uμ(x,0)u(x,0) weakly inL2(Ω). (99)

    Otherwise,

    um(x,0)u0(x) in H20(Ω). (100)

    Combining identities (99) and (100), we can get

    u(x,0)=u0(x).

    Next, according to (83), we can obtain

    (uμ,t,φk)(ut,φk) weakly * in L(0,T),
    (uμ,tt,φk)(utt,φk) weakly * in L(0,T).

    Then, we can discover

    (uμ,t(x,0),φk)(ut,φk)|t=0=(ut(x,0),φk). (101)

    Otherwise,

    (um,t(x,0),φk)(u1(x),φk), (102)

    Comparing identities (101) and (102), we can get

    (ut(x,0),φk)=(u1(x),φk), for arbitrary k.

    So

    ut(x,0)=u1(x).

    In the same way, we can obtain

    v(x,0)=v0(x),vt(x,0)=v1(x).

    Proof. Differentiating the first equation of (45) with respect to t, multiplying by utt and integrating with respect to x, we discover

    (uttt,utt)+((a6utxx)xx,utt)((a3vtxx)xx,utt)12(((u2x+v2x)utx)x,utt)((u2xutx)x,utt)((uxvxvtx)x,utt)2ωωt((Rux)x,utt)ω2((Rutx)x,utt)(p1,t,utt)=0. (103)

    Differentiating the second equation of (45) with respect to t, multiplying by vtt and integrating with respect to x, we discover

    (vttt,vtt)+((a4vtxx)xx,vtt)((a3utxx)xx,vtt)12(((u2x+v2x)vtx)x,vtt)((v2xvtx)x,vtt)((uxvxutx)x,vtt)2ωωt((Rvx)x,vtt)ω2((Rvtx)x,vtt)(p2,t,vtt)=0. (104)

    Summing the equation (103) and (104), we discover after integrating by parts:

    12ddt(||utt||2L2(Ω)+||vtt||2L2(Ω))+12ddt(||a6utxx||2L2(Ω)+||a4vtxx||2L2(Ω))ddt||a3utxxvtxx||L1(Ω)+ω22ddt||R(x)utx||2L2(Ω)+ω22ddt||R(x)vtx||2L2(Ω)=12((u2x+v2x)utxx,utt)+(u2xutxx,utt)+(uxvxvtxx,utt)+12((u2x+v2x)vtxx,vtt)+(v2xvtxx,vtt)+(uxvxutxx,vtt)+3(uxuxxutx,utt)+(vxvxxutx,utt)+(uxxvxvtx,utt)+(uxvxxvtx,utt)+(uxuxxvtx,vtt)+3(vxvxxvtx,vtt)+(uxxvxutx,vtt)+(uxvxxutx,vtt)+2ωωt((Rxux,utt)+(Ruxx,utt)+(Rxvx,vtt)+(Rvxx,vtt))+(p1,t,utt)+(p2,t,vtt). (105)

    Obviously, by Young inequality and Hölder inequality, we have

    12ddt(||utt||2L2(Ω)+||vtt||2L2(Ω))+12ddt(||a6utxx||2L2(Ω)+||a4vtxx||2L2(Ω))ddt||a3utxxvtxx||L(Ω)+ddt[ω22(||Rutx||2L2(Ω)+||Rvtx||2L2(Ω))]C(||ut||2H20(Ω)+||vt||2H20(Ω)+||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H20(Ω)+||v||2H20(Ω)+1). (106)

    Next integrate (106) with respect to t,

    ||utt||2L2(Ω)+||vtt||2L2(Ω)+||a6utxx||2L2(Ω)+||a4vtxx||2L2(Ω)2||a3utxxvtxx||L(Ω)+ω2(||Rutx||2L2(Ω)+||Rvtx||2L2(Ω))Ct0||ut||2H20(Ω)+||vt||2H20(Ω)+||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H20(Ω)+||v||2H20(Ω)dt+||utt(x,0)||2L2(Ω)+||utx(x,0)||2L2(Ω)+||utxx(x,0)||2L2(Ω)+||vtt(x,0)||2L2(Ω)+||vtx(x,0)||2L2(Ω)+||vtxx(x,0)||2L2(Ω)+C. (107)

    Since u1,v1H20(Ω), we have

    ||utxx(x,0)||L2(Ω),||vtxx(x,0)||L2(Ω),||utx(x,0)||L2(Ω),||vtx(x,0)||L2(Ω), (108)

    are bounded. On the other hand, multiplying the first equation and second equation of (45) by utt,vtt, respectively, and integrating with respect to x, we discover

    ||utt(x,0)||L2(Ω)C(||u0||H4(Ω)+||v0||H4(Ω))+CC,
    ||vtt(x,0)||L2(Ω)C(||u0||H4(Ω)+||v0||H4(Ω))+CC.

    where the condition (52), Young inequality and Poincaré inequality are used.

    As before, we have

    ||utt||2L2(Ω)+||vtt||2L2(Ω)+||ut||2H20(Ω)+||vt||2H20(Ω)Ct0(||utt||2L2(Ω)+||vtt||2L2(Ω)+||ut||2H20(Ω)+||vt||2H20(Ω))dt+C. (109)

    Applying the Gronwall inequality to (109) gives

    ut,vt   is bounded in   L(0,T;H20(Ω)),utt,vtt   is bounded in   L(0,T;L2(Ω)). (110)

    Moreover, the weak solutions of (45) satisfies

    utL(0,T;H20(Ω)),
    vtL(0,T;H20(Ω)),
    uttL(0,T;L2(Ω)),
    vttL(0,T;L2(Ω)).

    Proof. Multiplying the first equation and the second equation of (45) by φ2utxx, φ2vtxx, respectively, summing the two equations, and integrating with respect to x, we discover

    (utt,φ2utxx)+(vtt,φ2vtxx)+((a6uxx)xx,φ2utxx)+((a4vxx)xx,φ2vtxx)((a3vxx)xx,φ2utxx)((a3uxx)xx,φ2vtxx)12(((u2x+v2x)ux)x,φ2utxx)12(((u2x+v2x)vx)x,φ2vtxx)ω2((Rux)x,φ2utxx)ω2((Rvx)x,φ2vtxx)(p1,φ2utxx)(p2,φ2vtxx)=0. (111)

    Thanks to integration by parts and the properties of φ, we obtain the following equality

    I1+I2+I3+I4+I5=0. (112)

    where Ii,i=1,,5, are given as follows respectively

    I1=(utt,φ2utxx)+(vtt,φ2vtxx)I2=((a6uxx)xx,φ2utxx)+((a4vxx)xx,φ2vtxx)I3=((a3vxx)xx,φ2utxx)((a3uxx)xx,φ2vtxx)I4=12(((u2x+v2x)ux)x,φ2utxx)12(((u2x+v2x)vx)x,φ2vtxx)I5=(p1,φ2utxx)(p2,φ2vtxx).

    To conclude, we need to estimate each of Ii,i=1,,5. By using integration by parts and conclusion (53), we find

    I1=12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω))+2(φφxutt,utx)+2(φφxvtt,vtx)12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω))C (113)

    For I2, we get

    I2=12ddt(||a6φuxxx||2L2(Ω)+||a4φvxxx||2L2(Ω))+(a6,xuxx,φ2utxxx)+(a4,xvxx,φ2vtxxx)+2(a6uxxx,φφxutxx)+2(a4vxxx,φφxvtxx)+2(a6,xuxx,φφxutxx)+2(a4,xvxx,φφxvtxx)12ddt(||a6φuxxx||2L2(Ω)+||a4φvxxx||2L2(Ω))(a6,xuxxx,φ2utxx)(a4,xvxxx,φ2vtxx)+2(a6uxxx,φφxutxx)+2(a4vxxx,φφxvtxx)C12ddt(||a6φuxxx||2L2(Ω)+||a4φvxxx||2L2(Ω))C||φuxxx||2L2(Ω)C||φvxxx||2L2(Ω)C (114)

    where Lemma 3.1 and (52) are used. Similarly from (114), we deduce

    I3=(a3vxxx,φ2utxxx)(a3uxxx,φ2vtxxx)(a3,xvxx,φ2utxxx)(a3,xuxx,φ2vtxxx)2(a3vxxx,φφxutxx)2(a3uxxx,φφxvtxx)2(a3,xvxx,φφxutxx)2(a3,xuxx,φφxvtxx)=ddt||a3φ2uxxxvxxx||L1(Ω)(a3,xvxx,φ2utxxx)(a3,xuxx,φ2vtxxx)2(a3vxxx,φφxutxx)2(a3uxxx,φφxvtxx)2(a3,xvxx,φφxutxx)2(a3,xuxx,φφxvtxx)=ddt||a3φ2uxxxvxxx||L1(Ω)+(a3,xxvxx,φ2utxx)+(a3,xxuxx,φ2vtxx)+(a3,xvxxx,φφxutxx)+(a3,xuxxx,φφxvtxx)+2(a3,xvxx,φφxutxx)+2(a3,xuxx,φφxvtxx)2(a3vxxx,φφxutxx)2(a3uxxx,φφxvtxx)2(a3,xvxx,φφxutxx)2(a3,xuxx,φφxvtxx)ddt||a3φ2uxxxvxxx||L1(Ω)C||φuxxx||2L2(Ω)C||φvxxx||2L2(Ω)C (115)

    By using Lemma 3.1 and (52), we find

    I4=C||utxx||2L2(Ω)C||vtxx||2L2(Ω)C||uxx||2L2(Ω)c||vxx||2L2(Ω)C (116)
    I5=C(||ut||2H20(Ω)+||vt|2H20(Ω)+||u||2H20(Ω)+||v||2H20(Ω))CC (117)

    Putting (113)-(117) into (112), this yields

    12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω)+||a6φuxxx||2L2(Ω)+||a4φvxxx||2L2(Ω)        2||a3φ2uxxxvxxx||L1(Ω))C(||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω))+C (118)

    Integrating with respect to t(0,t), taking into account the condition (50), we discover

    ||φutx||2L2(Ω)+||φvtx||2L2(Ω)+||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω)
    Ct0||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω)dt+C(||u1||2H1(Ω)+||v1||2H1(Ω)+||u0||2H3(Ω)+||v0||2H3(Ω))+C (119)

    Thanks to the Gronwall inequality, we deduce

    ||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω)C (120)

    Proof. Utilizing to the first and second equations of (45), we can deduce

    ((a6a4a23)uxxxx,uxxxx)=(a4utta3vtt+ω2(R(a4ux+a3vx))x,uxxxx)+12(((u2x+v2x)(a4ux+a3vx))x,uxxxx)+((a4p1+a3p2),uxxxx). (121)

    Thanks to (50) and Hölder inequality, we see that

    ||uxxxx||L2(Ω)C(||utt+vtt+ut+vt||L2(Ω)+||u+v||H20(Ω))C. (122)

    Furthermore,

    ||uxxx||L2(Ω)C||uxxxx||47L2(Ω)||u||37L2(Ω)+C||u||L2(Ω)C, (123)

    where Gagliardo-Nirenberg inequality for bounded domains is used. Similarly, we deduce

    ||vxxx||L2(Ω)C,||vxxxx||L2(Ω)C. (124)

    Combining with the conclusions in Theorem (4.2) and Theorem (4.3), we can find that

    u,vL(0,T;H4(Ω)H20(Ω)).

    Proof. Denote

    {˜u,˜v}=π({˜a3,˜a4,˜a6,˜ω,˜θ,˜u0,˜v0,˜u1,˜v1}),
    η=u˜u,  ζ=v˜v.

    Then η,ζ satisfy

    12ddt(||ηt||2L2(Ω)+||ζt||2L2(Ω)+||a6ηxx||2L2(Ω)      +||a4ζxx||2L2(Ω)2||a3ηxxζxx||L(Ω))=12(((u2x+v2x)ux(˜u2x+˜v2x)˜ux)x,ηt)+12(((u2x+v2x)vx(˜u2x+˜v2x)˜vx)x,ζt)+ω2(((R(x)ηx)x,ηt)+((R(x)ζx)x,ζt))+ω2(η,ηt)+(¯p1,ηt)+(¯p2,ζt)
    ((a6˜a6)˜uxx,ηtxx)((a4˜a4)˜vxx,ζtxx)+((a3˜a3)˜vxx,ηtxx)+((a3˜a3)˜uxx,ζtxx)(ω2˜ω2)(R˜ux,ηtx)(ω2˜ω2)(R˜vx,ζtx)+(ω2˜ω2)(˜u,ηt)+(ωt˜ωt)(R0+x,ηt). (125)

    In equation (125), the nonlinear term satisfies

    ((u2x+v2x)ux(˜u2x+˜v2x)˜ux)x=((u2x+v2x)ηx)x+((u2x+v2x˜u2x˜v2x)˜ux)x=((u2x+v2x)ηx)x+(((ux+˜ux)ηx+(vx+˜vx)ζx)˜ux)x=2(uxuxx+vxvxx)ηx+(u2x+v2x)ηxx+((ux+˜ux)ηxx+(vx+˜vx)ζxx)˜ux+((uxx+˜uxx)ηx+(vxx+˜vxx)ζx)˜ux+((ux+˜ux)ηx+(vx+˜vx)ζx)˜uxx. (126)

    By Hölder inequality and Sobolev inequality, we can obtain

    (((u2x+v2x)ux(˜u2x+˜v2x)˜ux)x,ηt)C||uxx+vxx||L2(Ω)||ηx||L(Ω)||ηt||L2(Ω)+C||ηxx||L2(Ω)||ηt||L2(Ω)+||ηxx+(||uxx+˜uxx||L2(Ω)||ηx||L(Ω)+||vxx+˜vxx||L2(Ω)||ζx||L(Ω))||ηt||L2(Ω)+(|ηx||L(Ω)+||ζx||L(Ω))||˜uxx||L2(Ω)||ηt||L2(Ω)+ζxx||L2(Ω)||ηt||L2(Ω)C||η||H20(Ω)||ηt||L2(Ω)+C||η||H20(Ω)||ηt||L2(Ω)+C||η||H20(Ω)||ηt||L2(Ω)+C||ζ||H20(Ω)||ηt||L2(Ω)C(||ηt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)). (127)

    In the same way, we have

    (((u2x+v2x)vx(˜u2x+˜v2x)˜vx)x,ζt)C(||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)). (128)

    On the other hand, we easily have

    (¯p1,ηt)+(¯p2,ζt)C(||θ˜θ||L1(Ω)+||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H10(Ω)+||ζ||2H10(Ω)). (129)

    Substituting (127)-(129) into (125), we deduce

    ||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)Ct0||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)dt+||u1˜u1||2L2(Ω)+||v1˜v1||2L2(Ω)+||u0˜u0||2H20(Ω)+||v0˜v0||2H20(Ω)+C(5i=3||ai˜ai||L(Ω)+||θ˜θ||L1(Ω)+||ω˜ω||W1,1(0,T)). (130)

    where we used the inequalities

    ||˜u||L(Q)C,||˜ux||L(Q)C,||˜v||L(Q)C,||˜vx||L(Q)C,||˜uxx||L(0,T,L2(Ω))C,||˜vxx||L(0,T,L2(Ω))C,||ηtx||L(Q)C,||ζtx||L(Q)C,

    which are deduced from Theorem 4.2 and Theorem 4.3.

    Proof. Assume

    {˜φk}k=1is an orthogonal basis ofH2f(Ω), (131)

    and

    {˜φk}k=1is an orthonormal basis ofL2(Ω). (132)

    Fix a positive integer m, and write

    um=mk=1˜dk0m(t)˜φk, vm=mk=1˜dk1m(t)˜φk, (133)

    where we intend to select the coefficients ˜dk0m(t),˜dk1m(t)(0tT,k=1,,m) to satisfy

    ˜dk0m(0)=(u0,˜φk),˜dk1m(0)=(v0,˜φk),k=1,,m, (134)
    ˜dk0m,t(0)=(u1,˜φk),˜dk1m,t(0)=(v1,˜φk),k=1,,m, (135)

    and

    (um,tt,˜φk)+(a6um,xx,˜φk,xx)(a3vm,xx,˜φk,xx)+12((u2m,x+v2m,x)um,x,˜φk,x)+ω2(Rum,x,˜φk,x)(p1m,˜φk)(b1um(l)+e1vm(l))˜φ(l)=0, (136)
    (vm,tt,˜φk)(a3um,xx,˜φk,xx)+(a4vm,xx,˜φk,xx)+12((u2m,x+v2m,x)vm,x,˜φk,x)+ω2(Rvm,x,˜φk,x)(p2m,˜φk)(b2um(l)+e2vm(l))˜φ(l)=0. (137)

    As in earlier treatments of C-C boundary condition, we can conclude the following two conclude without difficulty.

    (ⅰ) For each integer m=1,2,, there exists a unique Galerkin approximations function um,vm of the from (133) satisfying (134)-(137) for 0tT.

    (ⅱ) u(x,0)=u0(x),v(x,0)=v0(x),ut(x,0)=u1(x),vt(x,0)=v1(x).

    Then, we proof the following estimate

    um,t2L2(Ω)+vm,t2L2(Ω)+um2H2f(Ω)+vm2H2f(Ω)C(u12L2(Ω)+v12L2(Ω)+u02H2f(Ω)+v02H2f(Ω))+C (138)

    Similarly from (75), we can deduce

    12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||a6um,xx||2L2(Ω)+||a4vm,xx||2L2(Ω)+ω22(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))}ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)12ddt{b1u2m(l)+e2v2m(l)+2e1um(l)vm(l)}=ωωt(||Rum,x||2L2(Ω)+||Rvm,x||2L2(Ω))+(p1m,um,t)+(p2m,vm,t), (139)

    where b2=e1 is used.

    Then we integrate (139) with respect to t, to discover

    ||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H2f(Ω)+||vm||2H2f(Ω)b1u2m(l)e2v2m(l)2e1um(l)vm(l)C(||um,t(0)||2L2(Ω)+||vm,t(0)||2L2(Ω)+||um(t)||2H2f(Ω)+||vm(t)||2H2f(Ω))b1u20m(l)e2v20m(l)2e1u0m(l)v0m(l)+Ct0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H2f(Ω)+||vm||2H2f(Ω)+C. (140)

    By simple calculation, we deduce

    b1u2m(l)e2v2m(l)2e1um(l)vm(l)0. (141)

    where b1e2e21 is used.

    On the other hand, according to u0m(x),v0m(x)H2f(Ω), by Sobolev inequality, we obtain

    u0m(x),v0m(x)C(¯Ω).

    Thus,

    u0m(l),v0m(l)C. (142)

    Substituting (141) and (142) into (140), applying Gronwall inequality, we can deduce (138).

    Now we pass to limits in our Galerkin approximations, applying estimate (138), we can discover (92), (93), (95). In order to complete the proof of the theorem, we just have to proof

    uμ(l)u(l),vμ(l)v(l), strongly in L(0,T). (143)

    where uμ,vμ are the convergent subsequence of um,vm, respectively.

    To verify this, recalling (138), we observe that

    uμu,vμv  strongly in  C(0,T;H1f(Ω)) (144)

    where the Corollary 4 of Chapter 8 in [30] is used.

    Furthermore, thanks to the conditions um(0)=u(0)=0, we obtain

    ||uμ(l)u(l)||L(0,T)=||(uμ(l)u(l))(uμ(0)u(0))||L(0,T)=||l0(uμ(x)u(x))xdx||L(0,T)l||(uμ(x)u(x))x||L(0,T;L2(Ω))l||uμ(x)u(x)||L(0,T;H1f(Ω)) (145)

    Thanks to (144), we can deduce

    uμ(l)u(l) strongly in L(0,T),

    Similarly, we have

    vμ(l)v(l) strongly in L(0,T).

    Proof. Similarly as (105), we can get

    12ddtE+12P1+P2=P3. (146)

    where

    E=||utt||2L2(Ω)+||vtt||2L2(Ω)+||a6utxx||2L2(Ω)+||a4vtxx||2L2(Ω)2||a3utxxvtxx||L(Ω)+ω2||R(x)utx||2L2(Ω)+ω2||R(x)vtx||2L2(Ω)
    P1=(((u2x+v2x)ux)t,uttx)+(((u2x+v2x)vx)t,vttx)P2=(b1u(l)+e1v(l))tutt(l)(b2u(l)+e2v(l))tvtt(l)P3=2ωωt(((Rux)x,utt)+((Rvx)x,vtt))+||Rutx||2L2(Ω)+||Rvtx||2L2(Ω))+(p1,t,utt)+(p2,t,vtt)

    By calculation, we can deduce

    P1=12ddt{||u2x+v2xutx||2L2(Ω)+||u2x+v2xvtx||2L2(Ω)+2||uxutx||2L2(Ω)              +2||vxvtx||2L2(Ω)+4||uxvxutxvtx||L1(Ω)}              3l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdx (147)
    P2=12ddt{b1u2t(l)+e2v2t(l)+2e1ut(l)vt(l)} (148)
    P3||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω)       +||utx||2L2(Ω)+||vtx||2L2(Ω) (149)

    Substituting (147), (148) and (149) into (146), we get

    12ddt{E+||u2x+v2xutx||2L2(Ω)+||u2x+v2xvtx||2L2(Ω)+2||uxutx||2L2(Ω)        +2||vxvtx||2L2(Ω)+4||uxvxutxvtx||L1(Ω)        (b1u2t(l)+e2v2t(l)+2e1ut(l)vt(l))}||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω)+||utx||2L2(Ω)+||vtx||2L2(Ω)        +3l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdx (150)

    Then integrate (150) with respect to t, to discover

    Et0||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω)    +||utx||2L2(Ω)+||vtx||2L2(Ω)dt    +3t0l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdxdt    +||utt(x,0)||2L2(Ω)+||u1||2H2f(Ω)+||vtt(x,0)||2L2(Ω)+||v1||2H2f(Ω).

    where

    b1u2t(l)e2v2t(l)2e1ut(l)vt(l)0

    and

    ||uxutx||2L2(Ω)+||vxvtx||2L2(Ω)+2||uxvxutxvtx||L1(Ω)0.

    are used.

    Similarly as (108), we discover

    ||utt(x,0)||L2(Ω),||vtt(x,0)||L2(Ω)c. (151)

    According to (151), we get

    Ect0||utt||2L2(Ω)+||vtt||2L2(Ω)+||utxx||2L2(Ω)+||vtxx||2L2(Ω)dt+3t0l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdxdt+c.

    By Hölder inequality and Sobolev inequality, we have

    Ect0{||utt||2L2(Ω)+||utxx||2L2(Ω)+||utxx||3L2(Ω)+||utxx||4L2(Ω)||vtt||2L2(Ω)+||vtxx||2L2(Ω)+||vtxx||3L2(Ω)+||vtxx||4L2(Ω)}dt+c. (152)

    Then, we obtain Theorem 5.3 by Gronwall inequality.



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